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Question #42998

Difficulty: 8/10 | 🟢 Free

The Question

A single-turn rectangular conducting loop is driven to rotate about its central axis of symmetry at a constant angular velocity $\omega$ . The loop consists of two rigid wires, each of length $l$ and mass $m$ , parallel to the rotation axis. The wires are connected by flexible, massless conducting cords of negligible tension. The two wires are mechanically constrained by a spring mechanism that exerts a radial restoring force of magnitude $k(r - r_0)$ on each wire directed towards the axis, where $r$ is the distance of the wire from the axis and $r_0$ is a positive constant. The loop has a total electrical resistance $R$ and self-inductance $L$ . It rotates in a uniform magnetic field of magnitude $B$ perpendicular to the rotation axis. The entire system is immersed in a viscous fluid that exerts a drag torque $\tau_d = b\omega$ on the loop. Assuming $k > m\omega^2$ , what is the average external mechanical power $P$ required to maintain the constant angular velocity $\omega$ ?

Options

A.
$\frac{2 B^2 k^2 r_0^2 l^2 \omega^2 R}{(R^2 + \omega^2 L^2)(k - m\omega^2)^2} + b\omega^2$
B.
$\frac{2 B^2 k^2 r_0^2 l^2 \omega^2}{R(k - m\omega^2)^2} + b\omega^2$
C.
$\frac{2 B^2 r_0^2 l^2 \omega^2 R}{(R^2 + \omega^2 L^2)} + b\omega^2$
D.
$\frac{2 B^2 k^2 r_0^2 l^2 \omega^2}{\sqrt{R^2 + \omega^2 L^2}(k - m\omega^2)^2} + b\omega^2$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

AC Generator Back-EMF|Viscosity and Stokes' Law|System of Particles and Rotational Motion

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of mechanics, AC circuit theory, and electromagnetic induction.

It features three distinct traps.

First, the 'Variable Rotational Inertia' trap: students often fail to account for the centrifugal expansion of the loop, incorrectly assuming the area remains constant at $A = 2r_0 l$ .

Second, the 'Inductive Power Factor' trap: students frequently ignore the self-inductance $L$ in an AC generator, assuming the electrical power is simply $\mathcal{E}_{rms}^2/R$ .

Third, even if they calculate the impedance $Z$ , they may forget that the current lags the EMF, erroneously using $I_{rms}^2 Z$ instead of $I_{rms}^2 R$ for the average power dissipation.

The correct approach requires coupling the mechanical equilibrium of the rotating frame with the power factor of an $RL$ circuit.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Electromagnetic Induction - Class 12
Concept	AC Generator Back-EMF
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free